Answer
$-3,-\frac{1}{2},\pm1$.
Work Step by Step
Step 1. For $f(x)=2x^4+7x^3+x^2-7x-3$, list possible rational zeros $\frac{p}{q}:\pm1,\pm3,\frac{1}{2},\pm\frac{3}{2}$.
Step 2. Use synthetic division as shown to find zero(s) $x=\pm1$.
Step 3. Use the quotient to find other zero(s) $2x^2+7x+3=0 \Longrightarrow (2x+1)(x+3)=0 \Longrightarrow x=-3,-\frac{1}{2}$.
